Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2001 AMC 10 Problems/Problem 22"

m (→‎Really Easy Solution)
Line 2: Line 2:
  
 
In the magic square shown, the sums of the numbers in each row, column, and diagonal are the same. Five of these numbers are represented by <math> v </math>, <math> w </math>, <math> x </math>, <math> y </math>, and <math> z </math>. Find <math> y + z </math>.
 
In the magic square shown, the sums of the numbers in each row, column, and diagonal are the same. Five of these numbers are represented by <math> v </math>, <math> w </math>, <math> x </math>, <math> y </math>, and <math> z </math>. Find <math> y + z </math>.
−
 
−
<math> \textbf{(A)}\ 43 \qquad \textbf{(B)}\ 44 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 46 \qquad \textbf{(E)}\ 47 </math>
 
  
 
<asy>
 
<asy>
Line 22: Line 20:
 
label("$24$",(1.5,2.5));
 
label("$24$",(1.5,2.5));
 
label("$w$",(2.5,2.5));</asy>
 
label("$w$",(2.5,2.5));</asy>
 +
 +
<math> \textbf{(A)}\ 43 \qquad \textbf{(B)}\ 44 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 46 \qquad \textbf{(E)}\ 47 </math>
 +
  
 
==Solutions==
 
==Solutions==

Revision as of 03:28, 13 January 2021

Problem

In the magic square shown, the sums of the numbers in each row, column, and diagonal are the same. Five of these numbers are represented by $v$, $w$, $x$, $y$, and $z$. Find $y + z$.

[asy] unitsize(10mm); defaultpen(linewidth(1pt)); for(int i=0; i<=3; ++i) { draw((0,i)--(3,i)); draw((i,0)--(i,3)); } label("$25$",(0.5,0.5)); label("$z$",(1.5,0.5)); label("$21$",(2.5,0.5)); label("$18$",(0.5,1.5)); label("$x$",(1.5,1.5)); label("$y$",(2.5,1.5)); label("$v$",(0.5,2.5)); label("$24$",(1.5,2.5)); label("$w$",(2.5,2.5));[/asy]

$\textbf{(A)}\ 43 \qquad \textbf{(B)}\ 44 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 46 \qquad \textbf{(E)}\ 47$


Solutions

Solution 1

We know that $y+z=2v$, so we could find one variable rather than two.

$v+24+w=43+v$

$24+w=43$

$w=19$

[asy] unitsize(10mm); defaultpen(linewidth(1pt)); for(int i=0; i<=3; ++i) { draw((0,i)--(3,i)); draw((i,0)--(i,3)); } label("$25$",(0.5,0.5)); label("$z$",(1.5,0.5)); label("$21$",(2.5,0.5)); label("$18$",(0.5,1.5)); label("$x$",(1.5,1.5)); label("$y$",(2.5,1.5)); label("$v$",(0.5,2.5)); label("$24$",(1.5,2.5)); label("$19$",(2.5,2.5));[/asy]

$44+x=24+x+z \implies z=20$

[asy] unitsize(10mm); defaultpen(linewidth(1pt)); for(int i=0; i<=3; ++i) { draw((0,i)--(3,i)); draw((i,0)--(i,3)); } label("$25$",(0.5,0.5)); label("$20$",(1.5,0.5)); label("$21$",(2.5,0.5)); label("$18$",(0.5,1.5)); label("$x$",(1.5,1.5)); label("$y$",(2.5,1.5)); label("$v$",(0.5,2.5)); label("$24$",(1.5,2.5)); label("$19$",(2.5,2.5));[/asy]

The sum per row is $25+21+20=66$.

Thus $66-18-25=66-43=v=23$.

Since we needed $2v$ and we know $v=23$, $23 \times 2 = \boxed{\textbf{(D)}\ 46}$.

Solution 2

$v+24+w=43+v$

$24+w=43$

$w=19$

[asy] unitsize(10mm); defaultpen(linewidth(1pt)); for(int i=0; i<=3; ++i) { draw((0,i)--(3,i)); draw((i,0)--(i,3)); } label("$25$",(0.5,0.5)); label("$z$",(1.5,0.5)); label("$21$",(2.5,0.5)); label("$18$",(0.5,1.5)); label("$x$",(1.5,1.5)); label("$y$",(2.5,1.5)); label("$v$",(0.5,2.5)); label("$24$",(1.5,2.5)); label("$19$",(2.5,2.5));[/asy]

$44+x=24+x+z \implies z=20$

[asy] unitsize(10mm); defaultpen(linewidth(1pt)); for(int i=0; i<=3; ++i) { draw((0,i)--(3,i)); draw((i,0)--(i,3)); } label("$25$",(0.5,0.5)); label("$20$",(1.5,0.5)); label("$21$",(2.5,0.5)); label("$18$",(0.5,1.5)); label("$x$",(1.5,1.5)); label("$y$",(2.5,1.5)); label("$v$",(0.5,2.5)); label("$24$",(1.5,2.5)); label("$19$",(2.5,2.5));[/asy]

The magic sum is determined by the bottom row. $25+20+21=66$.

Solving for $y$:

$y=66-19-21=66-40=26$.

To find our answer, we need to find $y+z$. $y+z=20+26 = \boxed{\textbf{(D)}\ 46}$.

Really Easy Solution

A nice thing to know is that any $3$ numbers that go through the middle form an arithmetic sequence.

Using this, we know that $x=(24+z)/2$, or $2x=24+z$ because $x$ would be the average.

We also know that because $x$ is the average the magic sum would be $3x$, so we can also write the equation $3x-46=z$ using the bottom row.

Solving for x in this system we get $x=22$, so now using the arithmetic sequence knowledge we find that $y=26$ and $z=20$.

Adding these we get $\boxed{\textbf{(D)}\ 46}$.


-harsha12345

Video Solution

https://youtu.be/9guPi81LgfM

~savannahsolver

See Also

2001 AMC 10 (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2001_AMC_10_Problems/Problem_22&oldid=142011"